1. 1 Unit Nine : Superposition Theorem John Elberfeld JElberfeld@itt-tech.edu WWW.J-Elberfeld.com ET115 DC Electronics

2. Schedule Unit Topic Chpt Labs 1.Quantities, Units, Safety12 (13) 2.Voltage, Current, Resistance23 + 16 3.Ohm’s Law35 (35) 4.Energy and Power36 (41) 5.Series CircuitsExam I47 (49) 6.Parallel Circuits59 (65) 7.Series-Parallel Circuits610 (75) 8.Thevenin’s, Power Exam 2619 (133) 9.Superposition Theorem 611 (81) 10.Magnetism & Magnetic Devices7Lab Final 11.Course Review and Final Exam 2

3. 3 Unit 9 Objectives - I State the superposition theorem.State the superposition theorem. List the steps required to apply the superposition theorem.List the steps required to apply the superposition theorem. Calculate the current and voltage in a given resistor by applying the superposition theorem.Calculate the current and voltage in a given resistor by applying the superposition theorem. Calculate the effect of a load on a bipolar voltage divided by applying the superposition theorem.Calculate the effect of a load on a bipolar voltage divided by applying the superposition theorem. Construct basic DC circuits on a protoboard.Construct basic DC circuits on a protoboard.

4. 4 Unit 9 Objectives – II Use a digital multimeter (DMM) to measure a predetermined low voltage on a power supply.Use a digital multimeter (DMM) to measure a predetermined low voltage on a power supply. Measure resistances and voltages in a DC circuit using a DMM.Measure resistances and voltages in a DC circuit using a DMM. Apply Ohm’s Law, Thevenin’s theorem, KVL and KCL to practical circuits.Apply Ohm’s Law, Thevenin’s theorem, KVL and KCL to practical circuits. Troubleshoot circuits constructed in Multisim exercises using simulated instrumentsTroubleshoot circuits constructed in Multisim exercises using simulated instruments

5. Reading Assignment Read and studyRead and study Chapter 6: Superposition Theorem: Pages 247-256 (Last section of chapter 6)Chapter 6: Superposition Theorem: Pages 247-256 (Last section of chapter 6) 5

6. Lab Assignment Experiment 11, “Superposition Theorem,” beginning on page 81 of DC Electronics: Lab Manual and MultiSim Guide.Experiment 11, “Superposition Theorem,” beginning on page 81 of DC Electronics: Lab Manual and MultiSim Guide. Complete all measurements, graphs, and questions and turn in your lab before leaving the roomComplete all measurements, graphs, and questions and turn in your lab before leaving the room 6

7. Written Assignments Complete the Unit 9 Homework sheetComplete the Unit 9 Homework sheet Show all your work!Show all your work! Be prepared for a quiz on questions similar to those on the homework.Be prepared for a quiz on questions similar to those on the homework. 7

8. 8 Superposition Theorem The current through or voltage drop across any component in a linear circuit with multiple current or voltage sources is equal to the algebraic sum of the currents or voltages produce by each source considered independentlyThe current through or voltage drop across any component in a linear circuit with multiple current or voltage sources is equal to the algebraic sum of the currents or voltages produce by each source considered independently

9. 9 Internal Resistances Actual batteries (Voltage sources) and current sources have an internal resistance that affects the circuit, in addition to voltage or current that they add to the circuitActual batteries (Voltage sources) and current sources have an internal resistance that affects the circuit, in addition to voltage or current that they add to the circuit Usually, we assume in internal resistance of a voltage source = 0, and the internal resistance of a current source is infinite (open)Usually, we assume in internal resistance of a voltage source = 0, and the internal resistance of a current source is infinite (open)

10. 10 Getting Equations Replace all but one of the voltage and current sources with internal resistancesReplace all but one of the voltage and current sources with internal resistances Find the current and voltages for the circuit for the one remaining sourceFind the current and voltages for the circuit for the one remaining source Repeat this process for each sourceRepeat this process for each source Find total voltages and currents by adding affects of each sourceFind total voltages and currents by adding affects of each source

11. 11Example Familiar setup of three resistors and two power suppliesFamiliar setup of three resistors and two power supplies Note the orientation of the power supplies!Note the orientation of the power supplies! V A2 = 40V R 3 =50Ω R 2 = 50Ω V A1 = 25V R 1 =75Ω VxVx

12. 12Example Replace V A2 with its internal resistance (0Ω)Replace V A2 with its internal resistance (0Ω) R 2||3 = 50Ω50Ω/(50Ω+50Ω) = 25 ΩR 2||3 = 50Ω50Ω/(50Ω+50Ω) = 25 Ω R 12||3 = 75 Ω + 25 Ω = 100ΩR 12||3 = 75 Ω + 25 Ω = 100Ω I T = 25V/100 Ω = 250ma = I R1I T = 25V/100 Ω = 250ma = I R1 V R1 = 250ma 75 Ω = 18.75VV R1 = 250ma 75 Ω = 18.75V V R2||3 =25V- 18.75V=6.25VV R2||3 =25V- 18.75V=6.25V I R2 =125maI R2 =125ma I R3 =125maI R3 =125ma V A2 = 40V R 3 =50Ω R 2 = 50Ω V A1 = 25V R 1 =75Ω

13. 13Example Replace V A1 with its internal resistance (0Ω)Replace V A1 with its internal resistance (0Ω) R 1||2 = 75Ω50Ω/(75Ω+50Ω) = 30 ΩR 1||2 = 75Ω50Ω/(75Ω+50Ω) = 30 Ω R 1||23 = 50 Ω + 30 Ω = 80ΩR 1||23 = 50 Ω + 30 Ω = 80Ω I T = 40V/80 Ω = 500ma = I R3I T = 40V/80 Ω = 500ma = I R3 V R3 = 500ma 50 Ω = 25VV R3 = 500ma 50 Ω = 25V V R1||2 =40V- 25V=15VV R1||2 =40V- 25V=15V I R1 =200maI R1 =200ma I R2 =300maI R2 =300ma V A2 = 40V R 3 =50Ω R 2 = 50Ω V A1 = 25V R 1 =75Ω

14. 14 Put It All Together V A2 = 40V R 3 =50Ω R 2 = 50Ω V A1 = 25V R 1 =75Ω 250ma, 18.75V 500ma, 25V 125ma, 6.25V 200ma, 15V 125ma, 6.25V 300ma, 15V

15. 15 Put It All Together V A2 = 40V R 3 =50Ω R 2 = 50Ω V A1 = 25V R 1 =75Ω 50ma, 3.75V 375ma, 18.75V 425ma, 21.25V

16. KCL and KVL Check: 425 mA = 50 mA + 375 A – OKCheck: 425 mA = 50 mA + 375 A – OK 25 V = 21.25 V + 3.75 V – OK25 V = 21.25 V + 3.75 V – OK 25 V = 3.75 V – 18.75 V + 40 V – OK25 V = 3.75 V – 18.75 V + 40 V – OK 40 V = 21.25 V + 18.75 V - OK40 V = 21.25 V + 18.75 V - OK 16 V A2 = 40V R 3 =50Ω R 2 = 50Ω V A1 = 25V R 1 =75Ω 50ma, 3.75V 375ma, 18.75V 425ma, 21.25V

17. 17 Superposition Theorem - Again This theorem states that if a linear network contains several independent energy sources, the total response is the sum of all the responses, if each source acted separately and all the other independent sources were replaced by their internal resistances.This theorem states that if a linear network contains several independent energy sources, the total response is the sum of all the responses, if each source acted separately and all the other independent sources were replaced by their internal resistances. This theorem is used to determine current in a network with multiple sources.This theorem is used to determine current in a network with multiple sources.

18. 18 Another Example In this example, the V A2 is flipped compared to previous examples!In this example, the V A2 is flipped compared to previous examples! V A2 = 10V R 3 =20Ω R 2 = 40Ω V A1 = 20V R 1 =10Ω

19. 19Example Replace V A2 with its internal resistance (0Ω)Replace V A2 with its internal resistance (0Ω) R 2||3 = 20Ω40Ω/(20Ω+40Ω) = 13.33 ΩR 2||3 = 20Ω40Ω/(20Ω+40Ω) = 13.33 Ω R 12||3 = 10 Ω + 13.33 Ω = 23.33ΩR 12||3 = 10 Ω + 13.33 Ω = 23.33Ω I T = 20V/23.33 Ω = 857ma = I R1I T = 20V/23.33 Ω = 857ma = I R1 V R1 = 857ma 10 Ω = 8.57VV R1 = 857ma 10 Ω = 8.57V V R2||3 =20V- 8.57V=11.43VV R2||3 =20V- 8.57V=11.43V I R2 =572maI R2 =572ma I R3 =285maI R3 =285ma R 3 =20Ω R 2 = 40Ω V A1 = 20V R 1 =10Ω

20. 20Example Replace V A1 with its internal resistance (0Ω)Replace V A1 with its internal resistance (0Ω) R 1||2 = 10Ω40Ω/(10Ω+40Ω) = 8 ΩR 1||2 = 10Ω40Ω/(10Ω+40Ω) = 8 Ω R 1||23 = 20 Ω + 8 Ω = 28ΩR 1||23 = 20 Ω + 8 Ω = 28Ω I T = 10V/28 Ω = 357ma = I R3I T = 10V/28 Ω = 357ma = I R3 V R3 = 357ma 20 Ω = 7.14VV R3 = 357ma 20 Ω = 7.14V V R1||2 =10V- 7.14V=2.86VV R1||2 =10V- 7.14V=2.86V I R1 =286maI R1 =286ma I R2 =71maI R2 =71ma V A2 = 10V R 3 =20Ω R 2 = 40Ω R 1 =10Ω

21. 21 Put It All Together 857ma, 8.57V 357ma, 7.14V 285ma, 11.43V 286ma, 2.86V 572ma, 11.43V 71ma, 2.86V V A2 = 10V R 3 =20Ω R 2 = 40Ω V A1 = 20V R 1 =10Ω

22. 22 KCL and KVL 1143ma, 11.43V 214ma, 8.57V 929ma, 18.57V V A2 = 10V R 3 =20Ω R 2 = 40Ω V A1 = 20V R 1 =10Ω 214 mA + 929 mA = 1143 mA – OK214 mA + 929 mA = 1143 mA – OK 20 V = 11.43 V + 8.75 V – OK20 V = 11.43 V + 8.75 V – OK 20 V = 11.43 V + 18.57 V – 10 V - OK20 V = 11.43 V + 18.57 V – 10 V - OK 10 V = 18,57 V – 8.57 V - OK10 V = 18,57 V – 8.57 V - OK

23. On your own… 23 V A2 = 10V R 3 =2 kΩ R 2 = 2 kΩ V A1 = 20V R 1 =4 kΩ

24. Short V 1 R 1 ||R 2 = 4kΩ 2kΩ/(4kΩ + 2kΩ) = 1.33kΩR 1 ||R 2 = 4kΩ 2kΩ/(4kΩ + 2kΩ) = 1.33kΩ R T = R 1 ||R 2 + R 3 = 2kΩ + 1.33kΩ = 3.33kΩR T = R 1 ||R 2 + R 3 = 2kΩ + 1.33kΩ = 3.33kΩ I T = 10V/ 3.33kΩ = 3 mAI T = 10V/ 3.33kΩ = 3 mA V R3 = I T R 3 = 3 mA 2kΩ = 6 VV R3 = I T R 3 = 3 mA 2kΩ = 6 V V R1||R2 = 10 V – 6 V = 4VV R1||R2 = 10 V – 6 V = 4V 24 V A2 = 10V R 3 =2 kΩ R 2 = 2 kΩ R 1 =4 kΩ

25. Short V 1 V R1 = 4 VI R1 = 4 V/ 4 kΩ = 1 mAV R1 = 4 VI R1 = 4 V/ 4 kΩ = 1 mA V R2 = 4 VI R2 = 4 V/ 2 kΩ = 2 mAV R2 = 4 VI R2 = 4 V/ 2 kΩ = 2 mA V R3 = 6 V I R3 = 3 mAV R3 = 6 V I R3 = 3 mA 25 V A2 = 10V R 3 =2 kΩ R 2 = 2 kΩ R 1 =4 kΩ

26. Short V 2 R 2 ||R 3 = 2kΩ 2kΩ/(2kΩ + 2kΩ) = 1 kΩR 2 ||R 3 = 2kΩ 2kΩ/(2kΩ + 2kΩ) = 1 kΩ R T = R 2 ||R 3 + R 2 = 2kΩ + 4 kΩ = 5 kΩR T = R 2 ||R 3 + R 2 = 2kΩ + 4 kΩ = 5 kΩ I T = 20V/ 5 kΩ = 4 mAI T = 20V/ 5 kΩ = 4 mA V R1 = I T R 1 = 4 mA 4 kΩ = 16 VV R1 = I T R 1 = 4 mA 4 kΩ = 16 V V R1||R2 = 20 V – 26 V = 4VV R1||R2 = 20 V – 26 V = 4V 26 V A1 = 20V R 3 =2 kΩ R 2 = 2 kΩ R 1 =4 kΩ

27. Short V 1 V R1 = 4 VI R1 = 4 V/ 2 kΩ = 2 mAV R1 = 4 VI R1 = 4 V/ 2 kΩ = 2 mA V R2 = 4 VI R2 = 4 V/ 2 kΩ = 2 mAV R2 = 4 VI R2 = 4 V/ 2 kΩ = 2 mA V R1 = 16 VI R1 = 4 mAV R1 = 16 VI R1 = 4 mA 27 V A2 = 10V R 3 =2 kΩ R 2 = 2 kΩ R 1 =4 kΩ

28. 28 Put It All Together 4 mA, 16 V 3 ma, 6 V 2 mA, 4 V 1 mA, 4 V 2 mA, 4 V V A2 = 10V R 3 =2kΩ R 2 = 2kΩ V A1 = 20V R 1 =4kΩ

29. 4 mA = 3 mA + 1 mA4 mA = 3 mA + 1 mA 20 V = 12 V + 8 V20 V = 12 V + 8 V 20 V = 12 V – 2 V + 10 V20 V = 12 V – 2 V + 10 V 10 V = 8 V + 2 V10 V = 8 V + 2 V 29 KCL and KVL 3 mA, 12 V 4 mA, 8 V 1 mA, 2 V V A2 = 10V R 3 =2 kΩ R 2 = 2 kΩ V A1 = 20V R 1 =4 kΩ

30. Summary Steps for the superposition theoremSteps for the superposition theorem Calculating multisource resistive circuits using superpositionCalculating multisource resistive circuits using superposition MultisimMultisim 30